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Romashka-Z-Leto [24]
3 years ago
15

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
5 0

Answer:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}

And, the distance from P to (0, -2) will be:

\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}

So sum of the two distances must be 6. Therefore:

d_1+d_2=6

Now, by substitution:

(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6

Simplify. We can subtract the second term from the left:

\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}

Square both sides:

(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)

We can cancel the x² terms and continue squaring:

y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

-8y=36-12\sqrt{x^2+(y+2)^2}

We can divide both sides by -4:

2y=-9+3\sqrt{x^2+(y+2)^2}

Adding 9 to both sides yields:

2y+9=3\sqrt{x^2+(y+2)^2}

And, we will square both sides one final time.

4y^2+36y+81=9(x^2+(y^2+4y+4))

Distribute:

4y^2+36y+81=9x^2+9y^2+36y+36

The 36y will cancel. So:

4y^2+81=9x^2+9y^2+36

Subtracting 4y² and 36 from both sides yields:

9x^2+5y^2=45

And dividing both sides by 45 produces:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

LUCKY_DIMON [66]3 years ago
4 0

ANSWER:

We need to know a distance formula to find the locus.

By distance formula,

  • √[x² + (y - 2)²] + √[x² + (y + 2)²] = 6.
  • √(x² + y² - 4y + 4) + √(x² + y² + 4y + 4) = 6.

Let x² + y² + 4 = a.

  • √(a - 4y) + √(a + 4y) = 6.
  • √(a - 4y) = 6 - √(a + 4y).

Squaring both sides,

  • a - 4y = a + 4y + 36 - 12 × √(a + 4y).
  • 8y + 36 = 12 × √(a + 4y).
  • 2y + 9 = 3 × √(a + 4y).

Squaring again,

  • 4y² + 36y + 81 = 9a + 36y.
  • 9a - 4y² - 81 = 0.

Putting back the original value of a,

  • 9x² + 5y² - 45 = 0.

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The sum of squares of numbers is: 13

Step-by-step explanation:

Let x and y be two numbers

Then,

Difference of the squares of the numbers will be:

x^2-y^2

Product will be:

xy

Given identity is:

(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2

Given values are:

Difference of the squares of the numbers=x^2-y^2=5

Product of numbers = xy = 6

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(x^2+y^2)^2=(5)^2+[2(6)]^2\\=25+(12)^2\\=25+144\\=169

As we have to only find x^2+y^2

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\sqrt{(x^2+y^2)^2}=\sqrt{169}\\x^2+y^2=13

The sum of squares of numbers is: 13

Keywords: Identities

Learn more about identities at:

  • brainly.com/question/8902155
  • brainly.com/question/8955867

#LearnwithBrainly

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